• sergiokapone
sergiokapone
If in the bottom of the bucket with water we made a hole, then water flows out of it. Bucket with water is the system with ##m(t)##, thus we have to use equation:
##m(t)\frac{dv}{dt} = F + u\frac{dm}{dt}##.
Where ##u\frac{dm}{dt}## -- is jet reaction force.
But, I think the jet reaction force should not appear in this case. What is the solution of this paradox?

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Gold Member
Hi, Sergio. If I understand the question correctly, there are two answers. A hole in the actually bottom would not provide thrust because the water would just be accelerating toward the ground as opposed to away from the bucket. That would lessen the effect of gravity upon the bucket because of weight loss, but that's not really a jet effect. If, on the other hand, the hole is in the side of the bucket near the bottom, there would be a tiny one because it would be pushing off from the water still in the container. I honestly think that I'm not correct in my interpretation of what you're asking though. Can you perhaps rephrase it?

Mentor
I don't understand the question.

sergiokapone
A hole in the actually bottom would not provide thrust because the water would just be accelerating toward the ground as opposed to away from the bucket. That would lessen the effect of gravity upon the bucket because of weight loss, but that's not really a jet effect.

I do not completely understand this. How I can write the equation of motion in the case the bucket, which is on the floor?

Gold Member
How I can write the equation of motion in the case the bucket, which is on the floor?
No help here... I don't know any math. I can say that there will be no motion at all if the thing is on the floor. (Unless you count immeasurably small structural vibrations on the molecular scale.) In fact, the water will just stop flowing completely if the surfaces of the bucket rim and the floor are smooth enough to form a seal.

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sergiokapone
It is strange to me, the in the case of the rocket the outgoing propellant create jet forse, but in the case of bucket - whater is not.

Gold Member
In the case of a rocket or air-breathing jet, or water jet or any other kind, the reaction mass is accelerated away from the container by either being expanded or being compressed and then released, or having the reaction chamber reduce in size to eject it (like in a syringe or balloon). In the bucket instance, gravity is simply pulling the water away from it. Earth is providing the force, not something in the bucket.

It is strange to me, the in the case of the rocket the outgoing propellant create jet forse, but in the case of bucket - whater is not.

This is how a rocket produces thrust:

Where is the green force in the case of your bucket?

sergiokapone
Earth is providing the force, not something in the bucket.

Ok, I understand that.

But now the quastion in the math. Why should I use equation of the motion ##m(t)\frac{dv}{dt}=F## instead ##m(t)\frac{dv}{dt}=F + u\frac{dm}{dt}##? Is the ##u## -equal zero in this case? But, obviously it is not.

But now the quastion in the math.
Define your symbols. What is F, v, m, u ?

Gold Member
Ok, I understand that.

Is the ##u## -equal zero in this case? But, obviously it is not.
As mentioned, I don't know any math at all beyond simple geometry. What is "u"? If it's physical displacement on an x, y, or z axis, it is zero.

sergiokapone
##u## - is the propellant relative velocity (relative to rocket or some other thing, bucket, for example e.t.c.)

##u## - is the relative velocity
Velocity of what and when? What are the other symbols representing?

sergiokapone
Define your symbols. What is F, v, m, u ?
##u## - is the propellant relative velocity
##F## - external force, gravity force, for example
##v## - is the velocity of the rocket
##m## - is the current mass of the rocket

Gold Member
##u## - is the propellant relative velocity (relative to rocket or some other thing, bucket, for example e.t.c.)
Okay, but I rather suspect that the equation should apply only if there is some physical interaction between the two objects. In this case, there isn't. That would be like using it to determine the force between an aeroplane and canoe. What applies in your case is the gravity formula and how the bucket becomes lighter as water leaves it.

edit: You guys are posting too quickly for me to keep up with. I'm going to take a bathroom/TV break (it's the only room with a flatscreen) and I'll check in later.

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Okay, that should work for you bucket too.
But, I think the jet reaction force should not appear in this case.
Why not? If you make the hole to the side, it will create thrust an propel the bucket horizontally.

sergiokapone
If you make the hole to the side, it will create thrust an propel the bucket horizontally.
This is not interesting trivial case.
What if I will create the hole in the bottom?

What if I will create the hole in the bottom?
Different direction of thrust.

Staff Emeritus
2022 Award
Hmm, a good question. My thoughts:

The reason that threre is no thrust is because there is no "top" of the bucket. If you poke a hole in the side, water still presses on the opposite side of the bucket and there is an imbalance of forces, which results in a net force on the bucket that makes it swing sideways.

However, since water isn't pressing on the top of the bucket, poking a hole on the bottom doesn't result in an imbalance of force from the water. Or, rather, there was already an imbalance of force since the water is pressing down but not up and we have simply lessened it.

Danger
The reason that threre is no thrust is because there is no "top" of the bucket.

But there is water on top of the water that is being pressed out of the bucket. The "thrust" (as defined in those equations) is not on the bucket itself, but on the water inside it. It reduces the vertical support force, that the bucket must apply to the water. This can also be attributed to the fact, that the water column above the hole is not supported by the bucket anymore.

Gold Member
the water column above the hole is not supported by the bucket anymore.
Which is exactly what I meant when I said that you are simply reducing the effect of gravity...

I swear that this is like talking to a wall...

Drakkith, your last post makes it about as clear as it can be. It's what I've been trying to say, but you said it better.

sergiokapone
Drakkith,
But my quastion is how looks the equation of variable-mass motion in this case and why its looks like?

sergiokapone
I need hard mathematical provement.

In the bucket instance, gravity is simply pulling the water away from it.
There is pressure in the bucket that presses the water out faster than just free fall from rest due to gravity pulling on it.

Earth is providing the force, not something in the bucket.
There is something in the bucket that provides a force: the water above the water that is being pressed out.

Gold Member
There is pressure in the bucket that presses the water out faster than just free fall from rest due to gravity pulling on it.

There is something in the bucket that provides a force: the water above the water that is being pressed out.
Both of which statements merely confirm what I've been saying; any impartation of force from the release is away from the water, not upon it and not upon the bucket. There is no physical transfer between the reservoir of water and the top of the bucket. The only force (other than the "negative" one) is upon the ground under the bucket.

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Homework Helper
For the hole on the bottom, the water flowing out of the hole experiences an acceleration due to the pressure at the bottom of the bucket (which is related to the density and height of the water in the bucket) and from the pull of gravity, so there's more than just gravity accelerating the water.

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Staff Emeritus
2022 Award
But there is water on top of the water that is being pressed out of the bucket. The "thrust" (as defined in those equations) is not on the bucket itself, but on the water inside it. It reduces the vertical support force, that the bucket must apply to the water. This can also be attributed to the fact, that the water column above the hole is not supported by the bucket anymore.

Hmm, I see. I wasn't aware that counted as "thrust".

Gold Member
Hmm, I see. I wasn't aware that counted as "thrust".
It doesn't, in any logical or physical sense of the term. (No wonder I hate math...)

edit: I wonder how sheepish NASA will feel when it's pointed out to them after all of these decades of spacecraft development that they could have just filled their ships with water and poked a hole in the bottom.

Mentor
I need hard mathematical provement.
The Wikipedia link you cited earlier includes the proof.

Hmm, I see. I wasn't aware that counted as "thrust".
It doesn't really matter how you call that term, the OP just asks whether it is zero in the bucket case. But the term "thrust" might be counter intuitive in the bottom hole bucket case, and the reason why the OP thinks it should be zero.

To make it more more intuitive think about a hole to the side, near the bottom. It clearly produces horizontal "thrust". Now attach a pipe to the hole that bends 90° downwards, and you have the same "thrust", but upwards, just like with the hole in the bucket bottom.

Staff Emeritus